3.132 \(\int \frac{\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=122 \[ -\frac{2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

[Out]

(a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3 + (b*(3*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d) -
b/(2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*a*b)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.212299, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3086, 3483, 3529, 3531, 3530} \[ -\frac{2 a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{a x \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(a*(a^2 - 3*b^2)*x)/(a^2 + b^2)^3 + (b*(3*a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d) -
b/(2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (2*a*b)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 3086

Int[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx &=\int \frac{1}{(a+b \tan (c+d x))^3} \, dx\\ &=-\frac{b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\int \frac{a-b \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx}{a^2+b^2}\\ &=-\frac{b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\int \frac{a^2-b^2-2 a b \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac{b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (b \left (3 a^2-b^2\right )\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=\frac{a \left (a^2-3 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac{b \left (3 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{b}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{2 a b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 1.15967, size = 154, normalized size = 1.26 \[ \frac{\frac{2 a \left (a^2-3 b^2\right ) (c+d x)}{\left (a^2+b^2\right )^3}+\frac{6 b^2 \sin (c+d x)}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac{2 b \left (b^2-3 a^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3}-\frac{b^3}{(a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((2*a*(a^2 - 3*b^2)*(c + d*x))/(a^2 + b^2)^3 - (2*b*(-3*a^2 + b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^2
+ b^2)^3 - b^3/((a - I*b)^2*(a + I*b)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + (6*b^2*Sin[c + d*x])/((a^2 + b^
2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])))/(2*d)

________________________________________________________________________________________

Maple [A]  time = 0.184, size = 219, normalized size = 1.8 \begin{align*} -{\frac{b}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-2\,{\frac{ab}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ){a}^{2}b}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ){b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

-1/2*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+3/d*b/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*a^2-1/d*b^3/(a^2+b^2)^3*ln(a+b*tan(
d*x+c))-2*a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))-3/2/d/(a^2+b^2)^3*ln(tan(d*x+c)^2+1)*a^2*b+1/2/d/(a^2+b^2)^3*ln(t
an(d*x+c)^2+1)*b^3+1/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^3-3/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a*b^2

________________________________________________________________________________________

Maxima [B]  time = 1.65487, size = 649, normalized size = 5.32 \begin{align*} \frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (-a - \frac{2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (\frac{{\left (3 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{{\left (5 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{{\left (3 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + \frac{4 \,{\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \,{\left (a^{8} - 3 \, a^{4} b^{4} - 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{4 \,{\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

(2*(a^3 - 3*a*b^2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^2*b - b^
3)*log(-a - 2*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^6 + 3*a^4*b^2 + 3*
a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6) + 2*((3*a^3*b^2 + a*b^4)*sin(d*x + c)/(cos(d*x + c) + 1) + (5*a^2*b^3 + b^5)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - (3*a^3*b^2 + a*b^4)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^8 + 2*a^6*b^2 + a^4*b^4 + 4*(a^7*b + 2*a^
5*b^3 + a^3*b^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(a^8 - 3*a^4*b^4 - 2*a^2*b^6)*sin(d*x + c)^2/(cos(d*x + c
) + 1)^2 - 4*(a^7*b + 2*a^5*b^3 + a^3*b^5)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + (a^8 + 2*a^6*b^2 + a^4*b^4)*s
in(d*x + c)^4/(cos(d*x + c) + 1)^4))/d

________________________________________________________________________________________

Fricas [B]  time = 0.547234, size = 748, normalized size = 6.13 \begin{align*} \frac{5 \, a^{2} b^{3} - b^{5} + 2 \,{\left (a^{3} b^{2} - 3 \, a b^{4}\right )} d x - 2 \,{\left (6 \, a^{2} b^{3} -{\left (a^{5} - 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b^{2} - 3 \, a b^{4} + 2 \,{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} d x\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (3 \, a^{2} b^{3} - b^{5} +{\left (3 \, a^{4} b - 4 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )}{2 \,{\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(5*a^2*b^3 - b^5 + 2*(a^3*b^2 - 3*a*b^4)*d*x - 2*(6*a^2*b^3 - (a^5 - 4*a^3*b^2 + 3*a*b^4)*d*x)*cos(d*x + c
)^2 + 2*(3*a^3*b^2 - 3*a*b^4 + 2*(a^4*b - 3*a^2*b^3)*d*x)*cos(d*x + c)*sin(d*x + c) + (3*a^2*b^3 - b^5 + (3*a^
4*b - 4*a^2*b^3 + b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - a*b^4)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c
)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2))/((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*
(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*
d)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [B]  time = 1.26319, size = 358, normalized size = 2.93 \begin{align*} \frac{\frac{2 \,{\left (a^{3} - 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (3 \, a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{9 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 3 \, b^{5} \tan \left (d x + c\right )^{2} + 22 \, a^{3} b^{2} \tan \left (d x + c\right ) - 2 \, a b^{4} \tan \left (d x + c\right ) + 14 \, a^{4} b + 3 \, a^{2} b^{3} + b^{5}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(a^3 - 3*a*b^2)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1)
/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b^2 - b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3
*a^2*b^5 + b^7) - (9*a^2*b^3*tan(d*x + c)^2 - 3*b^5*tan(d*x + c)^2 + 22*a^3*b^2*tan(d*x + c) - 2*a*b^4*tan(d*x
+ c) + 14*a^4*b + 3*a^2*b^3 + b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(b*tan(d*x + c) + a)^2))/d